7x^2+6x-99=-x^2-2x-3

Solution for 7x^2+6x-99=-x^2-2x-3 equation:

7x^2+6x-99=-x^2-2x-3

We move all terms to the left:

7x^2+6x-99-(-x^2-2x-3)=0

We get rid of parentheses

7x^2+x^2+2x+6x+3-99=0

We add all the numbers together, and all the variables

8x^2+8x-96=0

a = 8; b = 8; c = -96;
Δ = b2-4ac
Δ = 82-4·8·(-96)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-56}{2*8}=\frac{-64}{16}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+56}{2*8}=\frac{48}{16}
=3
$